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Problem 5.65
[Difficulty: 3]
Given:
Air flow through porous surface into narrow gap
Find:
The acceleration vector would be:
Solution:
We will apply the continuity equation to the control volume shown:
CS
CV
A
d
V
V
d
t
0
Governing
Equations:
(Continuity)
t
V
z
V
w
y
V
v
x
V
u
Dt
V
D
a
p
(Particle Accleration)
Assumptions:
(1) Steady flow
(2) Incompressible flow
(3) Uniform flow at every section
Based on the above assumptions the continuity equation reduces to:
0x
w
v
0
hw
ux
()
Solving for u:
() v
0
x
h
We apply the differential form of continuity to find v:
x
u
y
v
0
x
u
v
0
h
y
v
Therefore the yvelocity v is:
vy
v
0
h
d
fx
v
0
y
h
Now at y = 0:
vv
0
Therefore we can solve for f(x):
v
0
v
0
0
h
0
So we find that the ycomponent of velocity is:
0
1
y
h
y
V
v
x
V
u
a
p
Based on the above assumptions the particle acceleration reduces to:
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.
 Fall '07
 Lear
 Fluid Mechanics

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